I have a file called 'register.php' where person enters everything about their business plus sets username/password/confirmpass PLUS selects an image to upload ( a photo of themselves ). This all seems to work fine as all the data is being inserted properly (although it's not checking for dupes and allowing the same username to be added over and over..but that's another issue).
Now, I have page called 'members.php' that is to display their photograph and some other misc. crap. The key element is the photo. In the MySQL database in a table field is their photo's name. I have the query in the page to summon the table/field for it as so:
// Get Record Set $eg_recResult2 = mysql_query("SELECT `plateau_pros`.`egnID`, `plateau_pros`.`File1` FROM `plateau_pros` WHERE `plateau_pros`.`egnID` = '".@$eg_Result1['File1']."'", $eg_objConn1); $eg_Result2 = @mysql_fetch_array($eg_recResult2, MYSQL_ASSOC);
The code i'm trying to use for calling and displaying the image in the page has ranged from using an echo statement to whatever:
<?= @$eg_Result2['File1'] ?>
<? echo "<img src='http://www.plateauprofessionals.com/images/photo/$image' width='150' height='175'>";?>
using this variable set: $image = ($eg_Result2['File1']['name']);
I know this is basic stuff but honestly I can not find any tuts on a simple explanation on how to properly do this. Most want to splash a bunch of code up there and say 'and this is how we do it'. If I had some explanations on the steps for each part of the code then I would never have to ask this question again. Seriously, i've ready 100 image upload-to-mysql/dispaly-from-mysql tutorials and none of them explain it...they just show it.
Anyway, I get lightning fast answers here which makes it the hottest php resource i've encountered. Any help IS ALWAYS appreciated